Math Problem Statement

a cable 10 cm in diameter has a strength of 1018.6 MPa and the design engineer has specified a factor of safety of 2. what is the maximum allowable mass the cable can support?

Solution

To determine the maximum allowable mass the cable can support, we first need to follow these steps:

1. Convert the cable diameter into radius:

The diameter of the cable is given as 10 cm, so the radius rr is: r=102=5cm=0.05mr = \frac{10}{2} = 5 \, \text{cm} = 0.05 \, \text{m}

2. Calculate the cross-sectional area:

Assuming the cable has a circular cross-section, its area AA can be calculated using the formula for the area of a circle: A=πr2A = \pi r^2 Substituting r=0.05mr = 0.05 \, \text{m}: A=π(0.05)2=7.854×103m2A = \pi (0.05)^2 = 7.854 \times 10^{-3} \, \text{m}^2

3. Determine the allowable stress:

The allowable stress σallowable\sigma_{\text{allowable}} is the strength of the material divided by the factor of safety FSFS: σallowable=σmaterialFS=1018.6MPa2=509.3MPa\sigma_{\text{allowable}} = \frac{\sigma_{\text{material}}}{FS} = \frac{1018.6 \, \text{MPa}}{2} = 509.3 \, \text{MPa} Converting this into pascals (since 1MPa=106Pa1 \, \text{MPa} = 10^6 \, \text{Pa}): σallowable=509.3×106Pa\sigma_{\text{allowable}} = 509.3 \times 10^6 \, \text{Pa}

4. Calculate the maximum allowable force:

The maximum allowable force FmaxF_{\text{max}} is given by: Fmax=σallowable×AF_{\text{max}} = \sigma_{\text{allowable}} \times A Substituting the values for σallowable\sigma_{\text{allowable}} and AA: Fmax=(509.3×106)×(7.854×103)=4.000×106NF_{\text{max}} = (509.3 \times 10^6) \times (7.854 \times 10^{-3}) = 4.000 \times 10^6 \, \text{N}

5. Determine the maximum allowable mass:

The maximum allowable mass mmaxm_{\text{max}} is related to the maximum force by the equation: Fmax=mmax×gF_{\text{max}} = m_{\text{max}} \times g where gg is the acceleration due to gravity (approximately 9.81m/s29.81 \, \text{m/s}^2). Solving for mmaxm_{\text{max}}: mmax=Fmaxg=4.000×1069.81407,749.74kgm_{\text{max}} = \frac{F_{\text{max}}}{g} = \frac{4.000 \times 10^6}{9.81} \approx 407,749.74 \, \text{kg}

Conclusion:

The maximum allowable mass the cable can support is approximately 407,750 kg.

Would you like more details or have any questions?

5 Related Questions:

  1. How does the factor of safety affect the allowable stress in the cable?
  2. How would a different material strength change the maximum allowable mass?
  3. What would happen if the factor of safety was reduced?
  4. How can we design a cable to support more weight while keeping the same diameter?
  5. How would environmental factors, such as temperature, affect the cable’s strength?

Tip:

When designing with factors of safety, always ensure to consider dynamic loads and environmental conditions for maximum safety.

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Math Problem Analysis

Mathematical Concepts

Geometry (Circular Cross-Section)
Strength of Materials
Force and Mass Relationship
Safety Factor

Formulas

Area of a circle: A = πr^2
Allowable stress: σ_allowable = σ_material / FS
Force: F_max = σ_allowable × A
Mass: F_max = m_max × g

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Undergraduate Engineering